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just olya [345]
3 years ago
8

A student measures the mass and volume of a sample of aluminum at room temperature, and calculates the density of Aluminum to be

2.85 grams per cubic centimeter. Based on Table S, what is the percent error for the student’s calculated density of Aluminum?
Chemistry
2 answers:
erma4kov [3.2K]3 years ago
3 0

Answer:

From Table S you will take the value of S and then you will subtract (Table S Value-2.85) then divide by Table S value. After you will multiply by 100 to get percent.

Explanation:

efsa 2 years ago
0 0

5.6 is the answer to it

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What is extensibility?​
Katena32 [7]

Answer:

Extensibility is a software engineering and systems design principle that provides for future growth.

Explanation:

hope ot helps good day

5 0
3 years ago
Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0
3 years ago
Write the formula for lead oxide.
serious [3.7K]

Answer:

<h2>Lead(II) oxide</h2>

Explanation:

<h3>Lead(II) oxide, also called lead monoxide, is the inorganic compound with the molecular formula PbO. PbO occurs in two polymorphs: litharge having a tetragonal crystal structure, and massicot having an orthorhombic crystal structure. Modern applications for PbO are mostly in lead-based industrial glass and industrial ceramics, including computer components. It is an amphoteric oxide.[3]</h3>

  • IUPAC name
  • Lead(II) oxide

  • Other names
  • Lead monoxide
  • Litharge
  • Massicot
  • Plumbous oxide
  • Galena

<h2> Preparation</h2><h3>PbO may be prepared by heating lead metal in air at approximately 600 °C (1,100 °F). At this temperature it is also the end product of oxidation of other oxides of lead in air:[4]</h3><h3>Thermal decomposition of lead(II) nitrate or lead(II) carbonate also results in the formation of PbO:</h3>

<h3>2 Pb(NO</h3><h3>3)</h3><h3>2 → 2 PbO + 4 NO</h3><h3>2 + O</h3><h3>2</h3><h3>PbCO</h3><h3>3 → PbO + CO2</h3><h3>PbO is produced on a large scale as an intermediate product in refining raw lead ores into metallic lead. The usual lead ore is galena (lead(II) sulfide). At a temperature of around 1,000 °C (1,800 °F) the sulfide is converted to the oxide:[5]</h3>

<h3>2 PbS + 3 O</h3><h3>2 → 2 PbO + 2 SO2</h3><h3>Metallic lead is obtained by reducing PbO with carbon monoxide at around 1,200 °C (2,200 °F):[6]</h3>

<h3>PbO + CO → Pb + CO2</h3>

pls brainlest meh

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2 years ago
Please answer for me ☺️
levacccp [35]
They diffuse through small pores at the bottom of the leaf called stomata.

Plants can open and close their stomata, they open them to get water and carbon dioxide in but then close them to ensure the water doesn't 'leak' back out. This is important for plants growing in drier conditions like the cactus. 
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3 years ago
Question 6<br> How many molecules are in 4.48 liters of CO 2?
Dovator [93]
22.4 molecules are in 4.48 liters of CO 2
8 0
3 years ago
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